Question

In the arithmetic progression whose common difference is non-zero, the sum of first 3 n terms is equal to the sum of next n terms. Then the ratio of the sum of the first 2 n terms to the next 2 n terms is
(a) 1/5
(b) 2/3
(c) 3/4
(d) none of these

Answer 1

(a) 1/5

$$\begin{gathered} S_{3n}=S_{4n}-S_{3n} \\ \Rightarrow 2S_{3n}=S_{4n} \\ \Rightarrow 2\times \frac{3n}{2}\left\{2a+\left(3n-1\right)d\right\}=\frac{4n}{2}\left\{2a+\left(4n-1\right)d\right\} \\ \Rightarrow 3\left\{2a+\left(3n-1\right)d\right\}=2\left\{2a+\left(4n-1\right)d\right\} \\ \Rightarrow 6a+9nd-3d=4a+8nd-2d \\ \Rightarrow 2a+nd-d=0 \\ \Rightarrow 2a+\left(n-1\right)d=0....\left(1\right) \end{gathered}$$

Required ratio: $\frac{S_{2n}}{S_{4n}-S_{2n}}$

$$\begin{gathered} \frac{S_{2n}}{S_{4n}-S_{2n}}=\frac{{\displaystyle \frac{2n}{2}}\left\{2a+\left(2n-1\right)d\right\}}{{\displaystyle \frac{4n}{2}}\left\{2a+\left(4n-1\right)d\right\}-{\displaystyle \frac{2n}{2}}\left\{2a+\left(2n-1\right)d\right\}} \\ =\frac{n\left(nd\right)}{2n\left(3nd\right)-n\left(nd\right)} \\ =\frac{1}{6-1} \\ =\frac{1}{5} \end{gathered}$$